Can a function call be stored in a variable?

Let's see if I can explain what I would like to do (sorry for the inacuracy, I'm not a C expert).

I have a set of (similar) functions called depending on some value. Kind of (simplified):

Code:

---

int Func_0(int i)
{
...
    return j;
}
int Func_1(int i)
{
...
    return j;
}
...
int Func_n(int i)
{
...
    return j;
}

...
for (i=0; i<max_func; ++i)
{
    switch (i)
    {
      case 0:
            retval=Func_0(k);
            break;
      case 1:
            retval=Func_1(k);
            break;
      ...
      case n:
            retval=Func_n(k);
            break;
    }
    // use retval on some code
    ...
}

---

The real situation is more complex obviously, but I hope it's clear enough :blush:

Now, I would like to do something like ... (no laughing please! ;))

Code:

---

FuncType MyFuncs[max_func];                    // declare an array of "functions".
MyFuncs[0]=AddressOf(Func_0);    // store the function calls
MyFuncs[1]=AddressOf(Func_1);
...
MyFuncs[n]=AddressOf(Func_n);
...

// now use them

for (i=0; i<max_func; ++i)
{
    retval=MyFuncs[0](k);
    // use retval
  ...
}

---

I know most of it is not valid syntax, but I hope the idea is clear.

Is this possible?

Thks

What you are looking for is a "function pointer", pleanty of good reference links on on the topic already exist.....

You can use a function pointer.

Code:

---

int answer = 0;
//The functions
int sub(int n1, int n2);
int add(int n1, int n2);
int mult(int n1, int n2);
int div(int n1, int n2);
//The function pointer
int (*func1) (int , int);
//set the function pointer to the address of an arbitrary function
func1 = div;
//divide n1 by n2 with the function pointer and return the result
answer = func1(4,2);

---

Thks a lot. Works fine and saved me a lot of coding. :thumb: